Posted by **John** on Tuesday, March 26, 2013 at 11:25pm.

Consider all 2 digit numbers N such that the last 2 digits of N^2 are N itself. What is the sum of all these 2 digit numbers?

- Algebra -
**black_widow**, Wednesday, March 27, 2013 at 9:03am
the answer is 76

- Algebra -
**black_widow**, Wednesday, March 27, 2013 at 9:26am
sorry the answer should be 101

I do like this:

first consider the possible last digit of the number, then you have 0,1,5,6

Let N be ab

For number end with 0,

the last two digits of N will be 00

no numbers can be formed as 00 is not a

2 digit numbers

For number end with 1,

it must satisfy the conditions:

last digit of 2a=a

no numbers can be formed

For number end with 5,

it must satisfy the conditions:

last digit of 10a+2=a

the number 25 can be formed

For number end with ,

it must satisfy the conditions:

last digit of 12a+3=a

the number 76 can be formed

Sum of N=76+25=101

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